ID: A
UCSD Physics 2B
Answer Section
Unit Exam 1A
Charges, Force & Fields
MULTIPLE CHOICE
1. ANS: E
This is one of the basic facts about charges - from grade school, I think.
TOP: CHARGE CONCEPT
2. ANS: C
To apply the line/cylinder formula, we need the linear charge density λ . Pick some length L for a sample
section of cylinder. Then the lateral surface area of that section will be
A = 2π RL and the total charge is the area times the surface charge density Q = 2π RLσ .
Q
= 2π Rσ so that
L
2k λ 2k2π Rσ 4k π Rσ
E=
=
=
r
r
r
Now λ =
Ê
Ê
ˆÊ
ˆ
ˆ
4 ÁÁ 9.00 × 10 9 Nm2 / C 2 ˜˜ (3.14) ÁÁ 12.00 × 10−2 m˜˜ ÁÁ 14.0 × 10 −9 C / m2 ˜˜
Ë
N
kN
Ë
¯Ë
¯
¯
=
= 1,460 = 1.46
ÊÁ
−2 ˆ
C
C
Á 13.0 × 10 m˜˜
Ë
¯
The much easier way is to apply Gauss's Law from scratch:
Charged Cylinder Q = 2π RLσ
Gaussian Cylinder around the Charged Cylinder Area A = 2π rL
Q
2π RLσ
R σ
E=
=
=
ε 0 A 2π rLε 0 r ε 0
TOP: ELECTRIC FIELD CYLINDER
3. ANS: A
N
F
6.9 × 102 N
= 2.1 × 10 5
By definition, E = =
−3
C
q 3.3 × 10 C
TOP: ELECTRIC FIELD
1
ID: A
4. ANS: B
The electric field from a point charge is E =
kQ
r2
. Outside a charged spherical shell, the field looks the
same as if the whole charge were located in a point at the center. Halfway between the shells, the
3
R. Hence the field is
2
kQ
kQ
4 kQ
E= 2 =
=
2
9 R
ÊÁ 3
ˆ
r
ÁÁ R ˜˜˜
Ë 2 ¯
distance from their center is
Note that the field inside due to the outer charged shell is zero.
The other way to do this is to construct a Gaussian Surface at r = 3R / 2. The enclosed charge is +Q and
ÊÁ 3 ˆ˜ 2
the surface area A = 4π r 2 = 4π ÁÁÁÁ R ˜˜˜˜ = 9π R 2 . The electric field is
2
Q enc
E=
=
ε 0A
Ë
¯
+Q
1
4 kQ
ÊÁ 9 2 ˆ˜ = 9 R 2 where we used k = 4πε 0 .
4π ε 0 ÁÁÁÁ R ˜˜˜˜
Ë4 ¯
Again, Guass’ Law uses only the total charge enclosed.
TOP: ELECTRIC FIELD SPHERICAL
5. ANS: C
Make a ratio of the two forces using the fact that the only change is that the new distance is r new = 1.95 * r
k
F new
=
F
q1q2
ÊÁ r ˆ˜ 2
2
ˆ2
ÊÁ r ˆ˜ 2
Ê
Ë new ¯
˜˜ ⇒ F = F ÁÁÁ r ˜˜˜ = (33.0N ) ÁÊÁÁ 1 ˜ˆ˜˜ = 8.68N
= ÁÁÁÁ
˜˜
˜
Á
ÁÁ 1.95 ˜˜
new
ÁÁ r ˜˜
ÁË r new ˜¯
q1q2
¯
Ë
Ë new ¯
k
2
(r)
TOP: COULOMB FORCE CONCEPT
6. ANS: A
ˆ ÁÊ
−9 ˜
−9
ÁÊ
˜ˆ
2 ˆ
˜ ÁË 63.0 × 10 C ˜¯ ÁË −35.0 × 10 C ˜¯
q 1 q 2 ÊÁÁÁ
9 Nm ˜
˜
˜
F = k 2 = ÁÁÁ 9.00 × 10
= −3.18 × 10 −4 N
2 ˜
2
˜
Ê
ˆ
Á
˜
−2
r
C ¯
ÁÁ 25.0 × 10 m˜˜
Ë
Ë
¯
The minus sign shows that the force is attractive, as you’d expect between unlike charges.
TOP: COULOMB FORCE
2
ID: A
7. ANS: B
For a sphere with radial symmetry, we can take the volume element to be the product or the surface of a
spherical shell times its thickness dV = A surface × dr = 4π r 2 dr. We integrate the charge elements
dq = ρ (r)dV (r) as spherical shells since the density depends only on the radius r and is therefore the same
everywhere in each shell.
∫
∫
Q = ρ (r)dV = ρ (r)4π r 2 dr =
=
∫
R
0
ρo
4π ρ o
r
4π r 2 dr =
R
R
∫
R
r 3 dr
0
4π ρ o R 4
C ˜ˆ
Ê
ˆ 3 ÁÊ
= π R 3 ρ o = (3.14) ÁÁ 3.11 × 10 2 m˜˜ ÁÁÁÁ 2.97 × 10 −6 3 ˜˜˜˜ = 2.81 × 10 2 C
Ë
¯ Ë
R
4
m ¯
TOP: VOLUME CHARGE DENSITY
8. ANS: A
Ê
ˆÊ
ˆ
By definition, p = qd = ÁÁ 35 × 10 −9 C ˜˜ ÁÁ 8.0 × 10−2 m˜˜ = 2.8 × 10 −9 C ⋅ m
Ë
¯Ë
¯
The vector points towards the positive charge, i.e., down.
TOP: ELECTRIC DIPOLE
9. ANS: C
In case you don’t recall the formula (the hint given in Thursday lecture), just start with the formula for a single
kQ
point charge E = 2 since all point on the ring are equidistant from the point P. Then take the x-component (since
r
x
the others cancel by symmetry) and the factor is
r
kQ ÁÊÁ x ˜ˆ˜ kxQ
kxQ
kx (2π a λ )
E = 2 ÁÁÁ ˜˜˜ = 3 =
=
ÊÁ 2
ˆ3/2
ÊÁ 2
ˆ3/2
r
r Ër¯
Á x + a 2 ˜˜
Á x + a 2 ˜˜
Ë
¯
Ë
¯
ˆ
2 ˜
ÁÊÁ
Ê
ˆ
ÁÁ 9.00 × 10 9 Nm ˜˜˜ ÁÊ 36.0 × 10−2 m˜ˆ˜ 2 (3.14) ÁÊÁ 27.0 × 10 −2 m˜ˆ˜ ÁÁÁ 3.12 × 10 −4 C ˜˜˜
˜
ÁÁ
Á
Á
2 ˜Ë
m ˜˜¯
¯
¯ ÁË
Ë
˜
Á
C
¯
Ë
=
ÊÁ Ê
ˆ3/2
ÁÁ ÁÁ 36.0 × 10 −2 mˆ˜˜ 2 + ÊÁÁ 27.0 × 10 −2 mˆ˜˜ 2 ˜˜˜
ÁÁ Ë
¯
Ë
¯ ˜˜¯
Ë
= 1.88 × 10 7
N
C
TOP: ELECTRIC FIELD RING
10. ANS: C
ÁÊÁ
Á
charge
Q
6.60 × 10 −4 C
3 ÁÁÁ
ρ=
=
= ÁÁÁ
3
volume 4
4 ÁÁ
Ê
−2 ˆ
πR3
ÁÁ (3.14) ÁÁ 3.10 × 10 m˜˜
Ë
¯
3
Ë
˜ˆ˜
˜˜
˜˜
C
˜˜ = 5.29 3
˜˜
m
˜˜
˜¯
TOP: VOLUME CHARGE DENSITY
3
ID: A
11. ANS: B
This is a conversion problem, so use the identity e = 1.60 × 10−19 C to create "unit fraction"
ˆ˜
ÊÁ
total charge
˜˜
ÁÁ
1e +
ÊÁ
−9 ˆ
˜
˜ = 1.28 × 10 12
Á
#e =
= Á +205 × 10 C ˜ ÁÁ
¯ Á 1.60 × 10 −19 C ˜˜˜
charge/proton Ë
¯
Ë
TOP: ELEMENTARY CHARGE
12. ANS: C
N ˜ˆ
Ê
ˆ ÁÊ
τ = p × E = pE sin θ = ÁÁ 35 × 10 −9 C ⋅ m˜˜ ÁÁÁÁ 2.2 × 10 −3 ˜˜˜˜ sin165° = +2.0 × 10 −11 N ⋅ m
C¯
Ë
¯Ë
TOP: ELECTRIC DIPOLE TORQUE
13. ANS: E
The electric field inside any ideal conductor, even if it’s hollow & empty inside, is always zero.
TOP: CONDUCTOR FIELD CONCEPT
14. ANS: A
N ⋅ m2
ˆ
Ê
2
Φ E = E • A = EA cos θ = ÁÁ 31.7 × 10−6 N / C ˜˜ (2.66m) cos (11.6°) = 2.20 × 10 −4
¯
Ë
C
TOP: Electric Flux
15. ANS: B
The charge for each species is the number of particles times the charge per particle. The total net charge
is the sum of these:
Q net = n p q p + n e q e
Ê
ˆ
= ÁÁ n p e + ˜˜ + ÊÁÁ n e e − ˆ˜˜
¯
Ë
¯ Ë
Ê
ˆÊ
ˆÊ
ˆ Ê
ˆ
= ÁÁ 9.02 × 10 23 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ + ÁÁ 1.09 × 10 24 ˜˜ ÁÁ −1.60 × 10 −19 C ˜˜
Ë
¯Ë
¯Ë
¯ Ë
¯
Ê
ˆ
ˆÊ
= ÁÁ 9.02 × 10 23 − 1.09 × 1024 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜
Ë
Ë
¯
¯
ÊÁ
ˆ
Ê
ˆ
= Á −1.88 × 10 23 ˜˜ ÁÁ +1.60 × 10 −19 C ˜˜ = −3.01 × 10 4 C
Ë
¯Ë
¯
TOP: ELEMENTARY CHARGE
16. ANS: B
We apply Gauss’ Law by finding the total enclosed charge Q ENC = λ × circumference = λπ D
ΦE =
Q ENC
ε0
ÊÁ
ˆ
Ê
ˆ
Á 1.61 × 10 −6 C / m˜˜ (3.14) ÁÁ 26.3 × 10 −2 m˜˜
2
¯
Ë
¯
λπ D Ë
5 N ⋅m
=
=
=
1.50
×
10
ε0
C
8.85 × 10 −12 C 2 / N ⋅ m2
TOP: ELECTRIC FLUX
4
ID: A
17. ANS: D
For a single charged sheet, the field points perpendicular to the plane - away if positive and towards if
negative.
3.60 × 10 −9 C / m2
σ
N
E=
= Ê
= 2.03 × 10 2
2ε 0 2 ÁÁ 8.85 × 10−12 C 2 / N ⋅ m2 ˆ˜˜
C
Ë
¯
Note that the distance r doesn’t matter!
TOP: FIELD 1 SHEET
5